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2 years ago

7

If the frequency of an FM wave is 8.85 × 10¥ hertz, what is the period of the FM wave?

2 answers:

larisa [96]2 years ago

7 0

The value of the period of the wave is the reciprocal of its frequency in hertz (1/s). From the given above, the value of the period is calculated below,
t = 1 / f = 1 / (8.85 x 10^7 1/s)
Thus, the value of period (t) is equal to 1.13 x 10^-8 s. The answer is letter B.

natka813 [3]2 years ago

6 0

for plato users it is 1.10 X 10^-8 just took the test got 100 percent so ik its right

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2 years ago

What is the acceleration of a 59 kg object pushed with a force of 500 Newton's ?

Igoryamba

Answer:

a=500/50=10 m/s^2

Explanation:

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2 years ago

A car moving west slows down from a velocity of 50 m/s to 20 m/s in 15 seconds. What is its acceleration?

velikii [3]

Vf = Vo + at

Vf = 20 m/s
Vo = 50 m/s
a = ?
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2 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

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